Baekjun KOI Elementary School No. 14696 Ttakji Nori Conditional Question.

Hello, everyone Baekjun KOI Elementary School 14696 When it's a draw, I want to print out the result of D, but I think it's because of return, so how can I pick the result of the draw?

And I think my multi-conditional code is too complicated and I don't understand the part that goes down the conditional I wonder how I can modify it efficiently.

I'm referring to someone else's code, but I'm posting a question because I want to get feedback on the code I wrote. Thank you :)

# Enter
L = int(input()) # Total number of rounds
round = []
for i in range(2*L):
    A = [int(i) for i in list(map(int, input().split()))]
    round.append(A) # Cards of A and B for each round

# the number of the shape of the card
# # star = 4
# # circle = 3
# # square = 2
# # triangle = 1

defBattle_AB(round): # Count Comparison
    for i in range(0, L):
        If round[i][1:].count(4)!=round[i + 1][1:].count(4) # If the number of stars is different,
            if round[0][1:].count(4) > round[i + 1][1:].count(4):
                result = 'A'
                print(result)
            else:
                result = 'B'
                print(result)
        if round[i][1:].count(4) == round[i + 1][1:].count(4) # If the number of stars is the same,
            if round[i][1:].count(3) > round[i + 1][1:].count(3):
                result = 'A'
                print(result)
            elif round[i][1:].count(3) == round[i + 1][1:].count(3):
                return
            else:
                result = 'B'
                print(result)

                If round[i][1:].count(3) == round[i + 1][1:].count(3): # If the number of circles is the same,
                    if round[i][1:].count(2) > round[i + 1][1:].count(2):
                        result = 'A'
                    elif round[i][1:].count(2) == round[i + 1][1:].count(2):
                        return
                    else:
                        result = 'B'
                        print(result)

                    If round[i][1:].count(2) == round[i + 1][1:].count(2) # If the number of squares is the same,
                        if round[i][1:].count(1) > round[i + 1][1:].count(1):
                            result = 'A'
                            print(result)
                        elif round[i][1:].count(1) == round[i + 1][1:].count(1):
                            result = 'D'
                            print(result)
                        else:
                            result = 'B'
                            print(result)
        i += 2

    return result


# Execute
ab = Battle_AB(round)
print(ab)

coding-test

2022-09-20 16:53

1 Answers

First, look at the next development.

def getWinner(A, B) :

    if (number of stars varies):

        # At this point in time, actually, there's already a Shobu
        # We just need to figure out who won, return it, and end it

        별을urn 사람 r 더

    # When I came here, there was no Shobu because of the number of stars
    # Need to watch more

    if (the number of circles is different):
        The one who has more circles

    # The following is omitted

    # If you come all the way here, you can't tell who's the winner
    # Returns the default value at this time
    "Return" "No match"

The refactoring of the current code itself will not be difficult if the above development is understood.

So far, that's a boring answer, and let's think of something interesting. Even if you're in elementary school, if you're solving an engineering problem, when the code is getting complicated, Oh, isn't there a formula for this?You have to be able to think of .

Is there a formula for this problem? Let's try to induce him. According to the problem:

When I first saw this, I thought, "strong" is a concept of digits...It's called "

You have to give a card with 30 triangles on it so that you can't beat a card with a single star on it.

If you look at the problem conditions further, a ticket can hold up to 100 pictures, and no matter how many lower-class pictures there are, if a higher-class picture is not enough, you will be defeated unconditionally

I think it'll be about enough if I pay with an extra point, what do you think? Solve the task first, and think about it later!

2022-09-20 16:53

If you have any answers or tips

Popular Tags

python x 4647

android x 1593

java x 1494

javascript x 1427

c x 927

c++ x 878

ruby-on-rails x 696

php x 692

python3 x 685

html x 656